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# If n is a positive integer, the sum of the integers from 1 to n, inclu

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If n is a positive integer, the sum of the integers from 1 to n, inclu  [#permalink]

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Updated on: 30 Oct 2018, 01:12
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25% (medium)

Question Stats:

78% (01:18) correct 23% (01:37) wrong based on 43 sessions

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If n is a positive integer, the sum of the integers from 1 to n, inclusive, equals $$\frac{(n(n+1))}{2}$$ Which of the following equals the sum of the integers from 1 to 2n, inclusive?

A. $$n(n+1)$$

B. $$\frac{(n(2n+1))}{2}$$

C. $$n(2n+1)$$

D. $$2n(n+1)$$

E. $$2n(2n+1)$$

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Originally posted by pushpitkc on 30 Oct 2018, 01:08.
Last edited by Bunuel on 30 Oct 2018, 01:12, edited 1 time in total.
Edited the question.
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Re: If n is a positive integer, the sum of the integers from 1 to n, inclu  [#permalink]

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30 Oct 2018, 01:29
pushpitkc wrote:
If n is a positive integer, the sum of the integers from 1 to n, inclusive, equals $$\frac{(n(n+1))}{2}$$ Which of the following equals the sum of the integers from 1 to 2n, inclusive?

A. $$n(n+1)$$

B. $$\frac{(n(2n+1))}{2}$$

C. $$n(2n+1)$$

D. $$2n(n+1)$$

E. $$2n(2n+1)$$

sum of the integers from 1 to n, inclusive, equals $$\frac{(n(n+1))}{2}$$

Replacing $$n$$ by $$2n$$

sum of the integers from 1 to $$2n$$, inclusive, equals $$\frac{(2n(2n+1))}{2} = n(2n+1)$$

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Re: If n is a positive integer, the sum of the integers from 1 to n, inclu   [#permalink] 30 Oct 2018, 01:29
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